AMC10 2018 B

Answer (C): The numbers in the given sequence are of the form $10^n+1$ for $n=2,3,\ldots,2019$. If $n$ is even, say $n=2k$ for some positive integer $k$, then $10^n+1=100^k+1\equiv(-1)^k+1\pmod{101}$. Thus $10^n+1$ is divisible by $101$ if and only if $k$ is odd, which means $n=2,6,10,\ldots,2018$. There are $\frac14(2018-2)+1=505$ such values. On the other hand, if $n$ is odd, say $n=2k+1$ for some positive integer $k$, then $$ 10^n+1=10\cdot10^{n-1}+1=10\cdot100^k+1\equiv10\cdot(-1)^k+1\pmod{101}, $$ which is congruent to $9$ or $11$, and $10^n+1$ is not divisible by $101$ in this case.